Question: Simplify and expand the following expression: $ \dfrac{3}{t - 1}- \dfrac{3}{2t - 8}- \dfrac{2t}{t^2 - 5t + 4} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the second term: $ \dfrac{3}{2t - 8} = \dfrac{3}{2(t - 4)}$ We can factor the quadratic in the third term: $ \dfrac{2t}{t^2 - 5t + 4} = \dfrac{2t}{(t - 1)(t - 4)}$ Now we have: $ \dfrac{3}{t - 1}- \dfrac{3}{2(t - 4)}- \dfrac{2t}{(t - 1)(t - 4)} $ The least common multiple of the denominators is: $ (t - 1)(t - 4)$ In order to get the first term over $(t - 1)(t - 4)$ , multiply by $\dfrac{2(t - 4)}{2(t - 4)}$ $ \dfrac{3}{t - 1} \times \dfrac{2(t - 4)}{2(t - 4)} = \dfrac{6(t - 4)}{(t - 1)(t - 4)} $ In order to get the second term over $(t - 1)(t - 4)$ , multiply by $\dfrac{t - 1}{t - 1}$ $ \dfrac{3}{2(t - 4)} \times \dfrac{t - 1}{t - 1} = \dfrac{3(t - 1)}{(t - 1)(t - 4)} $ In order to get the third term over $(t - 1)(t - 4)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{2t}{(t - 1)(t - 4)} \times \dfrac{2}{2} = \dfrac{4t}{(t - 1)(t - 4)} $ Now we have: $ \dfrac{6(t - 4)}{(t - 1)(t - 4)} - \dfrac{3(t - 1)}{(t - 1)(t - 4)} - \dfrac{4t}{(t - 1)(t - 4)} $ $ = \dfrac{ 6(t - 4) - 3(t - 1) - 4t} {(t - 1)(t - 4)} $ Expand: $ = \dfrac{6t - 24 - 3t + 3 - 4t}{2t^2 - 10t + 8} $ $ = \dfrac{-t - 21}{2t^2 - 10t + 8}$